InterviewSolution
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InterviewSolution
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A gaseous mixture enclosed in a vessel of volume `V` consists of one gram mole of gas `A` with `gamma = (C_(P))/(C_(V)) = (5)/(3)`an another gas `B` with `gamma = (7)/(5)` at a certain temperature `T`. The gram molecular weights of the gases `A` and `B` are `4` and `32` respectively. The gases `A` and `B` do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation `PV^(19//13)` = constant , in adiabatic process. Find the number of gram moles of the gas `B` in the gaseous mixture.A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A Let the number of moles of gas B=n The number of moles of gas A=1 Since `U=(nRT)/(gamma-1)` For mixture, `U=U_(m),gamma=(19)/(13)` `therefore U_(m)=U_(A)+U_(B)` `therefore (n_(A)+n_(B)RT)/(gamma_(m)-1)=(n_(A)RT)/(gamma_(A)-1)+(n_(B)RT)/(gamma_(B)-1)` or `(1+n)/((19)/(13)-1)=(1)/((5)/(3)-1)+(n)/((7)/(5)-1)` or `(13(1+n))/(6)=(3)/(2)+(5n)/(2)or 13+13n=9+15n` or 4=2n or n=2 |
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